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\left(x-1\right)^{2}\geq 0
Divide both sides by 3. Since 3 is positive, the inequality direction remains the same. Zero divided by any non-zero number gives zero.
x^{2}-2x+1\geq 0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.
x^{2}-2x+1=0
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 1}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and 1 for c in the quadratic formula.
x=\frac{2±0}{2}
Do the calculations.
x=1
Solutions are the same.
\left(x-1\right)^{2}\geq 0
Rewrite the inequality by using the obtained solutions.
x\in \mathrm{R}
Inequality holds for x\in \mathrm{R}.